Answer:[tex]1.88\times 10^{14} V/m-s[/tex]
Explanation:
Given
radius of capacitor Plate [tex]r=2.8 cm[/tex]
Area [tex]A=\pi r^2=\pi \times 2.8^\times 10^{-4} m^2[/tex]
[tex]A=24.63\times 10^{-4} m^2[/tex]
current [tex]I=4.1 A[/tex]
separation [tex]d=1.1 mm[/tex]
Electric Field strength is given by
[tex]E=\frac{Q}{\epsilon _0A}[/tex]
[tex]E=\frac{I\cdot t}{\epsilon _0A}[/tex]
[tex]\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{I}{\epsilon _0A}[/tex]
[tex]\frac{\mathrm{d} E}{\mathrm{d} t}=\frac{4.1}{8.85\times 10^{-12}\times 24.63\times 10^{-4}}[/tex]
[tex]\frac{\mathrm{d} E}{\mathrm{d} t}=1.88\times 10^{14} V/m-s[/tex]
